Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#if 1

#include <bits/stdc++.h>
using namespace std;
double d[3005],p;
int n;

int main()
{
    d[0] = 1;
    scanf("%d", &n);
    for(int i = 0;i < n;i++)
    {
        scanf("%lf", &p);
        for(int j = i+1;j >= 1;j--)
        {
            d[j] += (d[j-1] - d[j]) * p;
        }
    }
    printf("%.10f",d[(n+1)/2]);
}

#endif // 1++